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# Stuck here, help me understand: Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

• Option 1)

0.025 x 104 m-1

• Option 2)

0.5 x 107 m-1

• Option 3)

0.25 x 107 m-1

• Option 4)

2.5 x 107 m-1

Answers (1)
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Balmer series -

$\frac{1}{\lambda }= R\left ( \frac{1}{2^{2}}- \frac{1}{n^{2}} \right )$

$\left ( \: visible \: light \right )$

- wherein

$n=3,4,5-----$

When electron jump from higher orbital to n=2 energy level

Balmer series $n\rightarrow 2$

last line of Balmer series means when $e^{-}$   jump from infinity $\rightarrow 2$

$\therefore \frac{1}{\lambda }= R\left ( \frac{1}{4}-\frac{1}{\infty } \right )=\frac{R}{4}$

Wave number $=\frac{1}{\lambda } = \frac{R}{4}$

$=0.25\times10^{7}\:m^{-1}$

Option 1)

0.025 x 104 m-1

This option is incorrect.

Option 2)

0.5 x 107 m-1

This option is incorrect.

Option 3)

0.25 x 107 m-1

This option is correct.

Option 4)

2.5 x 107 m-1

This option is incorrect.

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